When a mass $m$ is hung from the lower end of a spring of negligible mass,an extension $x$ is produced in the spring. The time period of oscillation is

$A$ light spring is suspended with mass $m_1$ at its lower end and its upper end is fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is $T$ seconds. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate, the period is found to be $\frac{3}{2} T$. The ratio $\frac{m_1}{m_2}$ is

$A$ spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio of $T_2 / T_1$ is

The total spring constant of the system as shown in the figure will be

The motion of a mass on a spring,with spring constant $K$ is as shown in the figure. The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{K}{m}}$. Suppose that at time $t = 0$,the position of the mass is $x(0)$ and velocity is $v(0)$,then its displacement can also be represented as $x(t) = C \cos (\omega t - \phi)$,where $C$ and $\phi$ are:

$A$ block of mass $M = 5 \ kg$ is suspended from one end of a spring. The spring extends by $l = 0.1 \ m$ due to the mass of the block. The block is given an upward velocity of $v = 2 \ m/s$. What is the maximum height $h$ (in $m$) reached by the block? $(g = 10 \ m/s^2)$